参考模拟退火程序：

#include <bits/stdc++.h>
using namespace std;
long double w[6][1012];
inline long double dis(long double a,long double b,long double c,long double d)
{
    return sqrtl((c-a)*(c-a)+(d-b)*(d-b));
}
inline bool blunt(long double a,long double b,long double c)
{
    return (a*a+c*c<b*b||a*a+b*b<c*c);
}
inline long double height(long double a,long double b,long double c)
{
    long double p=(a+b+c)/2;
    return 2*sqrtl(p*(p-a)*(p-b)*(p-c))/a;
}
inline long double check(long double dx,long double dy,int p)
{
    long double c=dis(dx,dy,w[0][p],w[1][p]);
    long double b=dis(dx,dy,w[2][p],w[3][p]);
    long double a=dis(w[0][p],w[1][p],w[2][p],w[3][p]);
    long double res=min(b,c);
    res=min(res,dis(dx,dy,w[4][p],w[5][p]));
    if(!blunt(a,b,c)) res=min(res,height(a,b,c));
    return res;
}
inline long double rd(long double m)
{
    return rand()*1.0/RAND_MAX*m;
}
int go()
{
    srand(time(0));
    int n;
    cin>>n;
    long double s=0;
    for(int i=1;i<=n;i++)
        for(int j=0;j<=5;j++)
        {
            cin>>w[j][i];
            s=max(s,w[j][i]);
        }
    long double S=pow(0.9999,-100000000.0/n),T=pow(0.9999,100000000.0/n);
    long double bx=s/2,by=s/2,bz=1e9;
    long double cx=s/2,cy=s/2,cz=1e9;
    long double f=1;
    while(S>T)
    {
        long double dx=bx+rd(s/10*f)-s/20*f;
        long double dy=by+rd(s/10*f)-s/20*f;
        long double dz=0;
        for(int i=1;i<=n;i++)
            dz=max(dz,check(dx,dy,i));
        if(dz<bz) bx=dx,by=dy,bz=dz,cx=dx,cy=dy,cz=dz;
        else if(rd(1)<exp((bz-dz)/S)) bx=dx,by=dy,bz=dz;
        f-=1e-8;
        S*=0.9999;
    }
    cout<<fixed<<setprecision(8)<<bx<<' '<<by<<' '<<bz<<endl;
    return 0;
}
int main()
{
    int T=1;
    //cin>>T;
    while(T--) go();
    return 0;
}

参考答案（$R = 4500.95088993$）：

5333.91419898 4918.20501141